By Peter M. Higgins

What are you able to do together with your maths? you should use it to completely comprehend all demeanour of items that can not be handled in the other manner. This ebook serves up numerous difficulties and exhibits how arithmetic solutions them. subject matters variety from cracking codes to the patience of recessive genes; from common sense puzzles to classical geometry; and from planetary movement inquiries to predicting the industry proportion of competing businesses. And there are different difficulties the place the maths itself is intrinsically incredible and interesting.

**Read Online or Download Professor Higgins’s Problem Collection PDF**

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**Extra resources for Professor Higgins’s Problem Collection**

**Sample text**

Here we say that, for a given positive integer n, two integers, a and b, are equal modulo n (more often we use the word congruent) if a – b = kn for some integer k. We write this as a ≡ b (mod n). For example, 20 ≡ 2 (mod 6) as 20 – 2 = 18 = 3 × 6. The arithmetic of the integers modulo n follows similar but not identical rules to that of ordinary algebra. We will not call upon the idea too much, except the notation crops up in the explanation of Problem 8 on coding theory. Combinations We will assume familiarity with expansions such as (a + b)2 = a2 + 2ab + b2 and the diﬀerence of two squares: a2 – b2 = (a – b)(a + b).

A famous inﬁnite series is the following: ∞ S= n=1 1 . n2 It comprises an inﬁnite sum of positive terms so it either diverges to +∞, as does for example 1 the sum of reciprocals, ∞ n=1 n , or converges to a limit. To prove that it converges to some limit, we only need to prove that the sum is bounded above by some positive number. This can be done in a number of ways, one of the quickest being the observation that the sum from n = 2 to ∞ is less than the area under the curve y = x12 from x = 1 to ∞.

1 so that (3) applies in all cases. Two consequences of these formulae are n n = k n–k 24 PROFESSOR HIGGINS’ S PROBLEM COLLECTION which follows from (3) and n 2n = k=0 n . k ( 4) To see the latter, put a = b = 1 in (2). Moreover, 2n is the total number of subsets of a set of size n, as the right-hand side of (4) sums the number of subsets of size k of a set of size n as k ranges from 0 up to n. Complex numbers Although not required in the previous problem set, so-called imaginary and complex numbers do arise later and, being numbers, we use this opportunity to outline their properties.